 A dumb circuit question.
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MikeT1982
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Sat Aug 04 2007, 12:36AM
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Registered Member #621
Joined: Sun Apr 01 2007, 12:37AM
Location:
Posts: 119
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This seems too obvious but bugs me. Say you have two 9 volt batteries and a light bulb. You connect one terminal of the light bulb to the Negative post of one battery, and the other terminal of the light bulb to the Positive terminal of the OTHER battery......But you DON'T jumper the remaining negative and Positive terminals together as you would to series the two batteries.
Of course the light bulb doesn't turn on as there is no current flow untill you jumper the remaining terminals and create a 18 volt potential.
Why though? If the one terminal has a negative potential and the other has a positive....why don't they "see" each other and allow current to flow even without the remaining two terminals being jumpered?
Mike
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Myke
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Registered Member #540
Joined: Mon Feb 19 2007, 07:49PM
Location: MIT
Posts: 969
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Your thinking of charges. The current can't flow unless there is a return path. It has to complete the circuit to work. In a tesla coil you may wonder why it will arc to your hand even though it isn't grounded. it is because you are capacitively coupled to ground. You could think of this as ground but with a cap.
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Sulaiman
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Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141
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Because each battery has a voltage difference between the terminals
the positive terminal is positive with respect to it's own negative terminal
neither terminal has a voltage with respect to ground/earth/universe
the battery terminals do not have an ABSOLUTE voltage, only a relative voltage
In your imaginary circuit,
connect the wire that goes to one of the battery negative terminals to 0v/ground/earth
The other terminal of that battery will be at +9V.
Because there is no current flowing through the bulb the positive terminal of the other battery will also be at 0v/ground/earth, so the negative terminal of this battery will be at -9V.
So if you put a volt meter across the two un-connected battery terminals
you will measure 18V.
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